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I cant figure out how to solve 4x+6<-6, and 
q+12-2(q-22)>0

asked Oct 26, 2013 in ALGEBRA 1 by homeworkhelp Mentor

2 Answers

0 votes

 1) Given ineqality 4x+6<-6

Apply subtraction property subtract each sides by 6

4x+6-6<-6-6

4x<-12

Apply division property of ineqality divide each sides by 4.

4x/4<-12/4

Cancel common on terms

x<-12/4

⇒x < -3

Graph of the inequality

2)Given ineqality q+12-2(q-22)>0 

q+12-2q+44>0( simplification)

q-2q+56>0

-q+56>0

-q>-56 (cancel same signs in ineqality)

q>56

Graph of ineqality

answered Oct 28, 2013 by william Mentor

2) Solution of the inequality q + 12 - 2(q - 22) > 0 is q < 56 .

0 votes

2) q + 12 - 2(q - 22) > 0

q + 12 - 2q + 44 > 0

-q + 56 > 0

Subtract 56 from each side.

-q + 56 - 56 > 0 - 56

-q > -56

Multiply each side by negitive one and flip the symbol of inequality.

q < 56

Solution {q| q < 56}

Solution on the number line.

q is less than 56,We must use open circle on 56,Since q is not equal to 56.

Then draw a line to the left to indicate that q can be any real number less than 56.

.

answered Jun 2, 2014 by david Expert

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